infinite product of interiors of sets in the box topology I know that in the product (Tychonoff) topology equality $${\rm int}\left(\prod_{\alpha \in I} A_
{\alpha}\right)= \prod_{\alpha \in I} {\rm int}\left( A_
{\alpha}\right)$$ does not hold true (see question Interior of arbitrary products). 
Does above equality hold in case the product space is given the box topology?
 A: Yes, it does hold.
Note that since each $\DeclareMathOperator{\Int}{Int}\Int (A_\alpha)$ is open in $X_\alpha$ it follows that $\prod_{\alpha \in I} \Int ( A_\alpha )$ is open in $\Box_{\alpha \in I} X_\alpha$, and is clearly a subset of $\prod_{\alpha \in I} A_\alpha$.  It thus follows that $${\textstyle \prod_{\alpha \in I}} \Int ( A_\alpha ) \subseteq \Int ( {\textstyle \prod_{\alpha \in I}} A_\alpha ).$$
Now let's take some $\langle x_\alpha \rangle_{\alpha \in I} \in \Int ( \prod_{\alpha \in I} A_\alpha )$.  But as $\Int ( \prod_{\alpha \in I} A_\alpha )$ is open, this means that for each $\alpha \in I$ there is an open $U_\alpha \subseteq X_\alpha$ containing $x_\alpha$ such that $${\textstyle \prod_{\alpha \in I}} U_\alpha \subseteq \Int ( {\textstyle \prod_{\alpha \in I}} A_\alpha ) \subseteq {\textstyle \prod_{\alpha \in I}} A_\alpha.$$ But in order for this to be true we must have that $U_\alpha \subseteq A_\alpha$ for all $\alpha$, which implies $U_\alpha \subseteq \Int ( A_\alpha)$.  Therefore $\langle x_\alpha \rangle_{\alpha \in I} \in \prod_{\alpha \in I} \Int ( A_\alpha )$.
